\documentclass{article}
\usepackage{amsmath,amssymb,amsthm}

% --- Page layout settings ---

% Set page margins
\usepackage[left=1.35in, right=1.35in, bottom=1in, top=1.1in, headsep=0.2in]{geometry}
% Anchor footnotes to the bottom of the page
\usepackage[bottom]{footmisc}

% Set line spacing
\renewcommand{\baselinestretch}{1.2}

% Set spacing between paragraphs
\setlength{\parskip}{1.5mm}

% Allow multi-line equations to break onto the next page
\allowdisplaybreaks

%math stuff%

% Numbered theorem, lemma, etc. settings - e.g., a definition, lemma, and theorem appearing in that 
% order in Section 2 will be numbered Definition 2.1, Lemma 2.2, Theorem 2.3. 
% Example usage: \begin{theorem}[Name of theorem] Theorem statement \end{theorem}

\theoremstyle{definition}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}

% Un-numbered theorem, lemma, etc. settings
% Example usage: \begin{lemma*}[Name of lemma] Lemma statement \end{lemma*}
\newtheorem*{theorem*}{Theorem}
\newtheorem*{proposition*}{Proposition}
\newtheorem*{lemma*}{Lemma}
\newtheorem*{corollary*}{Corollary}
\newtheorem*{definition*}{Definition}
\newtheorem*{example*}{Example}
\newtheorem*{remark*}{Remark}
\newtheorem*{claim}{Claim}


\begin{document}

\begin{section}*{Introduction}
 
\end{section}

\section{Lateral Surface}

A lateral surface $S$ is a region in $\mathbb{R}^3$ bounded by a curve $f \in \mathbb{R}^2$ that is revolved about a given axis, where the resulting surface $S \in \mathbb{R}^3$ bounds an object $f \in \mathbb{R}^3$ that is generated by the revolution of the curve $f \in \mathbb{R}^2$. 
	

\section{Lateral Surface Area}

The lateral surface area, $A_S$, is given by the area of $S \in \mathbb{R}^2$.


\section{Surface Area of Revolution}

A lateral surface $S \in \mathbb{R}^2$ may be rotated about an axis into $S \in \mathbb{R}^3$ by a trigonometric factor $T$ multiplied by $A_S$. Hence, for some solid of revolution, $R$ generated by $S \in \mathbb{R}^3$, the surface area of revolution is given by
$$A_R\ =\ TA_S.$$

\begin{example}
	$A_R = 2\pi A_S$, where the trigonometric factor $T = 2\pi$	imposes a full rotation of $2\pi$ radians upon $S \in \mathbb{R}^3$.
\end{example}

In some sense, the trigonometric factor may be viewed as a rotating function such that $T: \mathbb{R} \mapsto \mathbb{R}$. Alternately, $A_R$ is a function of rotation by $T: \mathbb{R} \mapsto \mathbb{R}$ such that $A_R: T\mapsto A_S$.

\begin{section}*{Computation}
 \section{Surface Area of Solids of Revolution}
 Given a smooth function $y = f(x)$ on the interval $[a, b]$ or $x = g(y)$ on the interval $[c, d]$, we can find the surface area of the solid generated by revolving the curve around the $x$-axis or the $y$-axis.
 
 \subsection{Functions of $x$}
 Let $y = f(x)$ be a smooth function on the interval $[a, b]$.
 
 \subsubsection{Revolution around the $x$-axis}
 The surface area $A$ of the solid generated by revolving the curve around the $x$-axis is given by
 \[
 	A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} dx.
 \]
 
 \textbf{Example:} Find the surface area of the solid generated by revolving the curve $y = \sqrt{1 - x^2}$, $0 \le x \le 1$, around the $x$-axis.
 \[
 	A = 2\pi \int_{0}^{1} \sqrt{1 - x^2} \sqrt{1 + \frac{x^2}{1 - x^2}} dx = 2\pi \int_{0}^{1} \sqrt{1 - x^2} \frac{1}{\sqrt{1 - x^2}} dx = 2\pi \int_{0}^{1} dx = 2\pi.
 \]
 
 \subsubsection{Revolution around the $y$-axis}
 The surface area $A$ of the solid generated by revolving the curve around the $y$-axis is given by
 \[
 	A = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx.
 \]
 
 \subsection{Functions of $y$}
 Let $x = g(y)$ be a smooth function on the interval $[c, d]$.
 
 \subsubsection{Revolution around the $y$-axis}
 The surface area $A$ of the solid generated by revolving the curve around the $y$-axis is given by
 \[
 	A = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy = 2\pi \int_{c}^{d} g(y) \sqrt{1 + \left(g'(y)\right)^2} dy.
 \]
 
 \textbf{Example:} Find the surface area of the solid generated by revolving the curve $x = \sqrt{1 - y^2}$, $0 \le y \le 1$, around the $y$-axis.
 \[
 	A = 2\pi \int_{0}^{1} \sqrt{1 - y^2} \sqrt{1 + \frac{y^2}{1 - y^2}} dy = 2\pi \int_{0}^{1}\sqrt{1 - y^2} \frac{1}{\sqrt{1 - y^2}} dy = 2\pi \int_{0}^{1} dy = 2\pi.
 \]
 
 \subsubsection{Revolution around the $x$-axis}
 The surface area $A$ of the solid generated by revolving the curve around the $x$-axis is given by
 \[
 	A = 2\pi \int_{c}^{d} y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy.
 \]
 
 \section{Summary}
 For a given smooth function $y = f(x)$ or $x = g(y)$, we can find the surface area of the solid generated by revolving the curve around the $x$-axis or $y$-axis using the following formulas:
 
 - For functions of $x$:
 - Revolution around the $x$-axis: $A = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} dx$.
 - Revolution around the $y$-axis: $A = 2\pi \int_{a}^{b} x \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx$.
 - For functions of $y$:
 - Revolution around the $y$-axis: $A = 2\pi \int_{c}^{d} g(y) \sqrt{1 + \left(g'(y)\right)^2} dy$.
 - Revolution around the $x$-axis: $A = 2\pi \int_{c}^{d} y \sqrt{1 + \left(\frac{dx}{dy}\right)^2} dy$.
 	
\end{section}

\end{document}